x→0
lim (arctanx)^ln(1-x)
=lim e^ ln (arctanx)^ln(1-x)
=e^lim ln (arctanx)^ln(1-x)
考慮
lim ln (arctanx)^ln(1-x)
=lim ln(1-x) * ln(arctanx)
根據(jù)等價無窮?。簂n(1+x)~x
=lim -x*ln(arctanx)
=-lim ln(arctanx) / 1/x
該極限為∞/∞型,根據(jù)L'Hospital法則
=-lim [ln(arctanx)]' / [1/x]'
=-lim 1/(arctanx)(1+x^2) / -1/x^2
=lim x^2 / (arctanx)(1+x^2)
根據(jù)等價無窮小：arctanx~x
=lim x / (1+x^2)
=0
故,原極限=e^0=1
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